CAIE Further Paper 1 2022 November — Question 6 15 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2022
SessionNovember
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeCommon perpendicular to two skew lines
DifficultyChallenging +1.3 This is a standard Further Maths vectors question on skew lines requiring the common perpendicular formula (using cross product and scalar triple product), followed by routine plane equations and angle calculations. While it involves multiple parts and several techniques, each step follows a well-established procedure taught explicitly in Further Maths courses, making it moderately above average difficulty but not requiring novel insight.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane4.04i Shortest distance: between a point and a line
6 The lines \(l _ { 1 }\) and \(l _ { 2 }\) have equations \(\mathbf { r } = 2 \mathbf { i } + \mathbf { k } + \lambda ( \mathbf { i } - \mathbf { j } + 2 \mathbf { k } )\) and \(\mathbf { r } = 2 \mathbf { j } + 6 \mathbf { k } + \mu ( \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k } )\) respectively. The point \(P\) on \(l _ { 1 }\) and the point \(Q\) on \(l _ { 2 }\) are such that \(P Q\) is perpendicular to both \(l _ { 1 }\) and \(l _ { 2 }\).
  1. Find the length \(P Q\).
    The plane \(\Pi _ { 1 }\) contains \(P Q\) and \(l _ { 1 }\).
    The plane \(\Pi _ { 2 }\) contains \(P Q\) and \(l _ { 2 }\).
    1. Write down an equation of \(\Pi _ { 1 }\), giving your answer in the form \(\mathbf { r } = \mathbf { a } + \mathbf { s b } + \mathbf { t c }\).
    2. Find an equation of \(\Pi _ { 2 }\), giving your answer in the form \(a x + b y + c z = d\).
  3. Find the acute angle between \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\).
Question 6(a):
AnswerMarks Guidance
\(\begin{pmatrix}0\\2\\6\end{pmatrix} - \begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}-2\\2\\5\end{pmatrix}\)B1 Finds a vector joining any point of \(l_1\) to any point of \(l_2\)
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 2 & -2\\1 & -1 & 2\end{vmatrix} = \begin{pmatrix}2\\-4\\-3\end{pmatrix}\)M1 A1 Finds common perpendicular
\(\frac{1}{\sqrt{29}}\left\begin{pmatrix}-2\\2\\5\end{pmatrix} \cdot \begin{pmatrix}2\\-4\\-3\end{pmatrix}\right = \frac{27}{\sqrt{29}} = 5.01\)
Total: 5
Question 6(b)(i):
AnswerMarks Guidance
\(\mathbf{r} = 2\mathbf{i} + \mathbf{k} + s(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + t(2\mathbf{i} - 4\mathbf{j} - 3\mathbf{k})\)B1 FT their common perpendicular from part (a)
Total: 1
Question 6(b)(ii):
AnswerMarks Guidance
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 2 & -2\\2 & -4 & -3\end{vmatrix} = \begin{pmatrix}-14\\-1\\-8\end{pmatrix}\)M1 A1FT Finds normal to plane \(\Pi_2\); FT their common perpendicular from part (a)
\(14(0) + (2) + 8(6) = 50 \Rightarrow 14x + y + 8z = 50\)M1 A1 Substitutes point
Total: 4
Question 6(c):
AnswerMarks Guidance
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & -1 & 2\\2 & -4 & -3\end{vmatrix} = \begin{pmatrix}11\\7\\-2\end{pmatrix}\)M1 A1 FT Finds normal to plane \(\Pi_1\); FT their part (b)(i)
\(\begin{pmatrix}11\\7\\-2\end{pmatrix} \cdot \begin{pmatrix}14\\1\\8\end{pmatrix} = \sqrt{174}\sqrt{261}\cos\theta\) leading to \(\cos\theta = \frac{145}{\sqrt{174}\sqrt{261}}\)M1 A1 Uses dot product of normal vectors
\(47.1°\)A1 0.822 radians
Total: 5
## Question 6(a):

$\begin{pmatrix}0\\2\\6\end{pmatrix} - \begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}-2\\2\\5\end{pmatrix}$ | B1 | Finds a vector joining any point of $l_1$ to any point of $l_2$

$\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 2 & -2\\1 & -1 & 2\end{vmatrix} = \begin{pmatrix}2\\-4\\-3\end{pmatrix}$ | M1 A1 | Finds common perpendicular

$\frac{1}{\sqrt{29}}\left|\begin{pmatrix}-2\\2\\5\end{pmatrix} \cdot \begin{pmatrix}2\\-4\\-3\end{pmatrix}\right| = \frac{27}{\sqrt{29}} = 5.01$ | M1 A1 | Uses formula for shortest distance

**Total: 5**

---

## Question 6(b)(i):

$\mathbf{r} = 2\mathbf{i} + \mathbf{k} + s(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + t(2\mathbf{i} - 4\mathbf{j} - 3\mathbf{k})$ | B1 | FT their common perpendicular from part (a)

**Total: 1**

---

## Question 6(b)(ii):

$\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 2 & -2\\2 & -4 & -3\end{vmatrix} = \begin{pmatrix}-14\\-1\\-8\end{pmatrix}$ | M1 A1FT | Finds normal to plane $\Pi_2$; FT their common perpendicular from part (a)

$14(0) + (2) + 8(6) = 50 \Rightarrow 14x + y + 8z = 50$ | M1 A1 | Substitutes point

**Total: 4**

---

## Question 6(c):

$\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & -1 & 2\\2 & -4 & -3\end{vmatrix} = \begin{pmatrix}11\\7\\-2\end{pmatrix}$ | M1 A1 FT | Finds normal to plane $\Pi_1$; FT their part (b)(i)

$\begin{pmatrix}11\\7\\-2\end{pmatrix} \cdot \begin{pmatrix}14\\1\\8\end{pmatrix} = \sqrt{174}\sqrt{261}\cos\theta$ leading to $\cos\theta = \frac{145}{\sqrt{174}\sqrt{261}}$ | M1 A1 | Uses dot product of normal vectors

$47.1°$ | A1 | 0.822 radians

**Total: 5**

---
6 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\mathbf { r } = 2 \mathbf { i } + \mathbf { k } + \lambda ( \mathbf { i } - \mathbf { j } + 2 \mathbf { k } )$ and $\mathbf { r } = 2 \mathbf { j } + 6 \mathbf { k } + \mu ( \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k } )$ respectively.

The point $P$ on $l _ { 1 }$ and the point $Q$ on $l _ { 2 }$ are such that $P Q$ is perpendicular to both $l _ { 1 }$ and $l _ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the length $P Q$.\\

The plane $\Pi _ { 1 }$ contains $P Q$ and $l _ { 1 }$.\\
The plane $\Pi _ { 2 }$ contains $P Q$ and $l _ { 2 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Write down an equation of $\Pi _ { 1 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \mathbf { s b } + \mathbf { t c }$.
\item Find an equation of $\Pi _ { 2 }$, giving your answer in the form $a x + b y + c z = d$.
\end{enumerate}\item Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q6 [15]}}
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